JEE Main 2019 — Coordination Compounds Question with Solution

CN of [Fe(H₂O)₆]Cl₂ = 6

For CN = 6, CFSE = $$\left[\frac{3}{5}\times n-\frac{2}{5}\times n_1\right]{\Delta }_0$$

Here, n = number of electron in e_g and
n₁ = number of electron in t_2g

Electronic configuration of Fe⁺² according to crystal field theory = $$t_{2g}^4{e}_g^2$$

$$\therefore$$ CFSE = $$\left[\frac{3}{5}\times 2-\frac{2}{5}\times 4\right]{\Delta }_0$$ = - 0.4 $${\Delta }_0$$

CN of K₂[NiCl₄] = 4

For CN = 4, CFSE = $$\left[\frac{2}{5}\times n-\frac{3}{5}\times n_1\right]{\Delta }_t$$

Here, n = number of electron in t_2g and
n₁ = number of electron in e_g

Electronic configuration of Ni⁺² according to crystal field theory = $$e_g^4{t}_{2g}^4$$

$$\therefore$$ CFSE = $$\left[\frac{2}{5}\times 4-\frac{3}{5}\times 4\right]{\Delta }_t$$ = - 0.8 $${\Delta }_t$$